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As the title already says it I have this expression $$ f(s)=\sum_{n=1}^\infty \frac{n^{s-1}}{e^{2\pi n}-1} $$ and am wondering if this one can be expressed in terms of any special or analytical functions?

Note that for large $n$ the denominator behaves like $e^{2\pi n}$ and thus approximately the main contributions come from the terms around the maximum $$n=\frac{s-1}{2\pi} $$ so that it should behave like $\sim (s-1)^{s-1}$.

In fact asymptotically it seems like $$(2\pi)^{s}f(s) \sim \Gamma(s) \, .$$

Thank you

Update: In writing $f(-s+1)$ I re-expressed $$n^{-s}= \frac{1}{\Gamma(s)}\int_{0}^\infty t^{s-1} \, e^{-nt} \, {\rm d}t \, ,$$ so that \begin{align} f(-s+1) &= \sum_{n=1}^\infty \frac{n^{-s}}{e^{2\pi n} -1} \\ &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}t \, t^{s-1} \sum_{n=1}^\infty e^{-nt} \sum_{k=0}^\infty e^{-2\pi n(k+1)} \\ &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}t \, t^{s-1} \sum_{k=0}^\infty \frac{1}{e^{t} \, e^{2\pi (k+1)} -1} \\ \end{align} and maybe $$\sum_{k=1}^\infty \frac{1}{z \, e^{2\pi k}-1}$$ is simpler to evaluate?

More than the precise values of $f(s)$ I'm actually interested in the analytic structure. Is it possible to see whether there are any zeros?

2nd Update: Being made aware about the Mellin-Transform Method shown at various sites here on MSE by Marko Riedel I carried out the analysis for certain positive $s$ values. For negative $s$ some more information can be found here: A new formula for Apery's constant and other zeta(s)?

We adapt the notation from above at little and write $$ f_k(x) = \sum_{n=1}^\infty \frac{n^{k-1}}{e^{nx}-1} $$ with Mellin-Transform $$ {\cal M}_{f_k}(s) = \Gamma(s)\zeta(s)\zeta(1+s-k) $$ where the integral converges for $s>k$. The inverse Mellin is given by $$ f_k(x)=\frac{1}{2\pi i} \int_{k+\frac{1}{2}-i\infty}^{k+\frac{1}{2}+i\infty} \Gamma(s)\zeta(s)\zeta(1+s-k) \, x^{-s} \, {\rm d}s \, . $$ We now close the contour at ${\rm Re}(s)=-\frac{1}{2}$ and transform this integral. First $$ \int_{-\frac{1}{2}+i\infty}^{-\frac{1}{2}-i\infty} \Gamma(s)\zeta(s)\zeta(1+s-k) \, x^{-s} \, {\rm d}s \\ \stackrel{u=k-s}{=} -\int_{k+\frac{1}{2}-i\infty}^{k+\frac{1}{2}+i\infty} \Gamma(k-u)\zeta(k-u)\zeta(1-u) \, x^{u-k} \, {\rm d}u $$ and then using the $\zeta$-functional equation $$ \zeta(1-s) = \frac{2}{\left(2\pi\right)^s} \, \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s) $$ and continue (renaming $u$ back to $s$) $$ =(-1)^{\frac{k}{2}+1}\int_{k+\frac{1}{2}-i\infty}^{k+\frac{1}{2}+i\infty} \Gamma(s)\zeta(s)\zeta(1+s-k) \, \left(2\pi\right)^{-s} \, \left(\frac{x}{2\pi}\right)^{s-k} \, {\rm d}s $$ where we needed to assume $k$ even. Obviously the value $x=2\pi$ is somewhat special and plugging in this becomes $$ =(-1)^{\frac{k}{2}+1} \, 2\pi i \, f_k\left(2\pi\right) \, . $$

The poles we enclosed are at $s=k$, $s=1$ and $s=0$ and the residue theorem yields $$ \left(1-\left(-1\right)^{\frac{k}{2}}\right) f_k(2\pi) = \frac{1}{2\pi i} \oint_\gamma \Gamma(s)\zeta(s) \zeta(1+s-k) \, (2\pi)^{-s} \, {\rm d}s \\ = \frac{\Gamma(k)\zeta(k)}{(2\pi)^k} + \frac{\Gamma(1)\zeta(2-k)}{2\pi} + \zeta(0)\zeta(1-k) \, . $$ In this equation not all even $k$ values are allowed, but only those of the form $$ k=4n+2 \qquad n=0,1,2,3,... $$ and we obtain the result $$ f_{4n+2}(2\pi) = \frac{(4n+2)!\zeta(4n+2)}{2(4n+2)(2\pi)^{4n+2}} + \frac{\zeta(-4n)}{4\pi} - \frac{\zeta(-4n-1)}{4} \\ = \frac{{\rm B}_{4n+2}}{2(4n+2)} - \frac{\delta_{n,0}}{8\pi} \, . $$

In fact the latter result also seems to be very close for the numbers $$ k=4n+4 \qquad n=0,1,2,3,... $$ if we correct the negative sign of these Bernoulli numbers $$ f_{2n}(2\pi) = \frac{(-1)^{n+1} \, {\rm B}_{2n}}{2(2n)} - \frac{\delta_{n,1}}{8\pi} \qquad n=1,2,3,... \, . $$

Diger
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    I don't believe that this is a representation of a special function. – Mark Viola Aug 15 '18 at 03:14
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    When $s$ is even, there are special values in terms of $\Gamma(1/4)$. – pisco Aug 15 '18 at 04:16
  • @pisco, How do you see these special values for $s$ even in terms of $\Gamma(1/4)$? Is there already a formula? – Diger Sep 02 '18 at 11:57
  • @Diger There is a way to find them out systematically, although an explicit formula might be quite complicated. – pisco Sep 03 '18 at 10:39
  • I don't know if you had another look at my update, but for values of the form $4n+2$ the result is just related to the Bernoulli-Numbers.

    So what do you mean by find out systematically ? By which approach ? Can you give an example ?

     – Diger Sep 03 '18 at 11:35
  • You may have a look at this answer https://math.stackexchange.com/a/1944103/72031 – Paramanand Singh Sep 24 '18 at 16:24
  • Hey, thanks i finished reading the first part. Is there also a pdf file for the second? https://paramanands.blogspot.com/2013/05/certain-lambert-series-identities-and-their-proof-via-trigonometry-part-2.html#.W6pNjRq1Kf0 – Diger Sep 25 '18 at 15:04
  • Yes there is a pdf link at the end of second post also. You can get pdf at https://sites.google.com/site/parama/Certain%20Lambert%20Series%20Identities%20and%20their%20Proof%20via%20Trigonometry_Part%202.pdf Also do try to open the blog posts with browser in desktop mode. They are not designed for mobile yet. – Paramanand Singh Sep 27 '18 at 08:07

2 Answers2

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As Mark Viola commented, I do not think that any special function will represent $f(s)$.

What we can notice is that the $-1$ in denominator is almost negligible compared to $e^{2n \pi}$ even for $n=1$ $(e^{2\pi} \approx 535.492)$ and then $$f(s)\approx\sum_{n=1}^\infty \frac{n^{s-1}}{e^{2\pi n}}=\text{Li}_{1-s}\left(e^{-2 \pi }\right)$$ where appears to polylogarithm function. The error is quite marginal as shown below $$\left( \begin{array}{cc} s & 100 \,\frac{ f(s)-\text{Li}_{1-s}\left(e^{-2 \pi }\right)}{f(s)}\\ 0 & 0.18657 \\ 1 & 0.18640 \\ 2 & 0.18605 \\ 3 & 0.18536 \\ 4 & 0.18399 \\ 5 & 0.18130 \\ 6 & 0.17611 \\ 7 & 0.16650 \\ 8 & 0.14990 \\ 9 & 0.12457 \\ 10 & 0.09240 \\ 11 & 0.06005 \\ 12 & 0.03443 \\ 13 & 0.01787 \\ 14 & 0.00862 \\ 15 & 0.00395 \\ 16 & 0.00174 \\ 17 & 0.00076 \\ 18 & 0.00033 \\ 19 & 0.00015 \\ 20 & 0.00008 \end{array} \right)$$

Edit

We can improve the result writing $$\frac 1 {e^{2n \pi}-1}=\sum_{k=1}^\infty e^{-2kn \pi}$$ and get $$f(s)=\sum_{k=1}^\infty \text{Li}_{1-s}\left(e^{-2k \pi }\right)$$ what could be approximated as $$f(s)=\sum_{k=1}^p\text{Li}_{1-s}\left(e^{-2k \pi }\right)+\sum_{k=p+1}^\infty \text{Li}_{1-s}\left(e^{-2k \pi }\right)$$ and use for the last summation the asymptotic $$\text{Li}_{1-s}\left(x\right)=x+2^{s-1} x^2+3^{s-1} x^3+O\left(x^4\right)$$ Using $p=2$ already leads to extremely good values.

Claude Leibovici
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  • @Diger, just an idea, did you try to use that $\zeta(s)\Gamma(s)=\int_0^\infty \frac {x^{s-1}}{(e^{x} - 1) }dx$ (https://en.wikipedia.org/wiki/Gamma_function) ? – JV.Stalker Aug 15 '18 at 07:36
  • Do you mean to approximate? I don't see how the integral is just the sum as I was looking for an exact expression. Or do you mean by using Euler-Maclaurin summation formula: $$ \sum_{i = m}^{\infty} a(i,s) = \int_{m}^{\infty} a(x,s) ,\mathrm{d}x + \frac{a(\infty,s) + a(m,s)}{2} + \sum_{j = 1}^{\infty} \frac{B_{2j}}{(2j)!} \left( a^{(2j-1)}(\infty,s) - a^{(2j-1)}(m,s) \right) $$ where $a(n,s)=\frac{n^{s-1}}{e^{2\pi n}-1}$ ? – Diger Aug 15 '18 at 10:15
  • Yes, it could be some kind of approximation. – JV.Stalker Aug 15 '18 at 10:20
  • As it looks like, the derivatives of $a$ at $m=1$ i.e. $$\frac{a^{(2j-1)}(1,s)}{(2\pi)^{2j}}$$ for the estimate of the remainder term are blowing up though. – Diger Aug 15 '18 at 11:02
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These series are related to modular forms, the Dedekind eta function, in particular. Ramanujan, for instance, proved that $f(14)=\frac{1}{24}$ (have a look at page 68 of my notes). The computation is related to the Poisson summation formula. To prove $f(2)=\frac{1}{24}-\frac{1}{8\pi}$ is fairly simple and for negative values of $s$ a standard technique is the Mellin transform; it is nicely explained here by Marko Riedel. A relation with double series of the $\sum_{m\geq 1}\sum_{n\geq 1}\frac{1}{m^{2\tau}(m^2+n^2)}$ kind is outlined here.

Jack D'Aurizio
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  • Hey, in your notes there is this sum $$\sum_{n=1}^\infty \frac{1}{\sinh^2(\pi n)} = \frac{1}{6} - \frac{1}{2\pi}$$. How do you prove this? – Diger Aug 15 '18 at 23:56
  • In fact $$ f(2)=\frac{1}{4} \sum_{n=1}^\infty \frac{1}{\sinh^2(\pi n)} , .$$ – Diger Aug 16 '18 at 09:34
  • Using your hints I only get $$ \sum_{n=1}^\infty \frac{1}{\sinh^2 (\pi n)} \ -\frac{1}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \frac{{\rm d}^2}{{\rm d}z^2} \left{ \sum_{k=1}^\infty \log\left(1+\frac{n^2z^2}{k^2}\right) + \log(\pi n z) \right} \Bigg|{z=1} \ =- \frac{2}{\pi^2} \sum{n=1}^\infty \sum_{k=1}^\infty \frac{k^2-n^2}{(k^2+n^2)^2} + \frac{1}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \ =-\frac{1}{\pi^2} \sum_{n=1}^\infty \sum_{k=-\infty}^\infty \frac{k^2-n^2}{(k^2+n^2)^2} $$ – Diger Aug 16 '18 at 10:25
  • $$=-\frac{1}{\pi^2} \sum_{n=1}^\infty \sum_{k=-\infty}^\infty \int_{-\infty}^\infty \frac{t^2-n^2}{(t^2+n^2)^2} , e^{-2\pi i kt} , {\rm d}t \ =4 \sum_{n=1}^\infty \sum_{k=1}^\infty k , e^{-2\pi k n} \ =\sum_{n=1}^\infty \frac{1}{\sinh^2(\pi n)} , . $$ – Diger Aug 16 '18 at 10:25
  • Do you see the issue I have? – Diger Sep 01 '18 at 16:56
  • @Diger: if you do not exchange any sums it is not surprising that you just turn the starting series into itself. $$\sum_{n,k\geq 1}\frac{k}{e^{2\pi k n}}=\sum_{k\geq 1}\frac{k}{e^{2\pi k}-1}$$ can be easily evaluated by residues, as shown by Marko Riedel. – Jack D'Aurizio Sep 01 '18 at 17:01
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    Ok sure. I just thought your hints implied it could also be shown in a different way, not by residues and by following these hints I just wanted to illustrate that I arrive at the beginning. – Diger Sep 01 '18 at 17:10